Firstly, I thought E
given a,b,c,d are positive integers. Is c divisible by a/d implies \(\frac{c}{a/d}=\frac{cd}{a}=int\)
(1) b * c is divisible by a
Factor 'a' can be present in 'b' or in 'c'..... We cannot exactly say if 'c' has factor 'a' to make \(\frac{cd}{a}\) integer because 'b' can also have a factor 'a' or both 'b', 'c' can have a factor of 'a'- Not Sufficient
(2) GCF (a,b) = d
GCF of a,b is d in the sense..... both 'a' and 'b' will have a common factor 'd'.... but you do not know exactly what is 'a'... 'a' can just be 'd' or '2d' or '3d' or 'nd' in which case \(\frac{cd}{a}\) has a remaining factor from 'a' (except 'a'='d' case all other cases have factors such as 2,3, or n) in the denominator..... For \(\frac{cd}{a}\) to be an integer the remaining factor of 'a' should also be present in 'c'. Not sufficient.
(1) + (2)
there are still possible cases of both things to happen...
My working was as below.
GCF (a,b) = d. In other words we can write as \(a=Ad\) and \(b=Bd\) where (A, B) should be (even, odd) or (odd, even) or (prime numbers).
From 1, \(c=integer*a/b\) implies \(c = integer*A/B\)
Now consider \(\frac{cd}{a}\). substituting values \(\frac{cd}{a}\)= \(integer*A/B *d/Ad\) = \(integer/B\). Since there are no restrictions with Integer and B, I went with E.... that fraction can be an integer or a non-integer.
But I am not able to create examples for showing \(\frac{cd}{a}\) is not integer are yielding contradictory values.... Answer may be C.... Can someone decode the logic from where I left off. Thanks.